如何通過擬合R中的線性b樣條回歸來提取基礎系數？

Question

例如，以下一級結，樣條曲線：

library(splines)
library(ISLR)

age.grid    = seq(range(Wage$age)[1], range(Wage$age)[2])
fit.spline  = lm(wage~bs(age, knots=c(30), degree=1), data=Wage)
pred.spline = predict(fit.spline, newdata=list(age=age.grid), se=T)

plot(Wage$age, Wage$wage, col="gray")
lines(age.grid, pred.spline$fit, col="red")

# NOTE: This is **NOT** the same as fitting two piece-wise linear models becase
# the spline will add the contraint that the function is continuous at age=30
# fit.1  = lm(wage~age, data=subset(Wage,age<30))
# fit.2  = lm(wage~age, data=subset(Wage,age>=30))

有沒有一種方法可以提取結之前和之后的線性模型（及其系數）？ 也就是說，如何在age=30切入點age=30之前和之后提取兩個線性模型？

使用summary(fit.spline)產生系數，但是（據我所知）它們對於解釋沒有意義。

Answer 1

您可以像這樣從fit.spline手動提取系數

summary(fit.spline)

Call:
lm(formula = wage ~ bs(age, knots = 30, degree = 1), data = Wage)
Coefficients:
                                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)                         54.19       4.05    13.4   <2e-16 ***
bs(age, knots = 30, degree = 1)1    58.43       4.61    12.7   <2e-16 ***
bs(age, knots = 30, degree = 1)2    68.73       4.54    15.1   <2e-16 ***
---

range(Wage$age)
## [1] 18 80
## coefficients of the first model
a1 <- seq(18, 30, length.out = 10)
b1 <- seq(54.19, 58.43+54.19, length.out = 10)
## coefficients of the second model
a2 <- seq(30, 80, length.out = 10)
b2 <- seq(54.19 + 58.43, 54.19 + 68.73, length.out = 10)
plot(Wage$age, Wage$wage, col="gray", xlim = c(0, 90))
lines(x = a1, y = b1, col = "blue" )
lines(x = a2, y = b2, col = "red")

如果您想要像線性模型一樣的斜率系數，則可以簡單地使用

b1 <- (58.43)/(30 - 18)
b2 <- (68.73 - 58.43)/(80 - 30)

請注意，在fit.spline ，截距是指age = 18時的wage值，而在線性模型中，截距是指age = 0時的wage值。

Answer 2

當您在bspline回歸中預先指定自由度時，主要是完成提取結的操作。 例：

fit.spline = lm（工資〜bs（年齡，df = 5），數據=工資）

attr（bs（age，df = 5），“結”）

33.33333％66.66667％

  37 48 

可以在第293頁的ISLR書（您似乎正在使用）中找到一個示例。